What is the correct order of reactivity of alcohols toward reactions involving cleavage of C-O bond?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3° > 2° > 1°
Reactions that break the C–O bond of an alcohol, such as the reaction with HCl/\( \ce{ZnCl2} \), go through a carbocation. The –OH group is first protonated and then lost as water:
\( \displaystyle \ce{R-OH + H+ <=> R-OH2+ -> R+ + H2O} \)
A carbocation is more stable the more alkyl groups it carries (+I effect and hyperconjugation), so the C–O bond breaks most easily where the most stable cation forms:
3° (first, most reactive) — gives a tertiary carbocation, stabilised by three alkyl groups; it reacts with concentrated HCl even at room temperature.
2° (second) — gives a secondary carbocation with two alkyl groups; it needs \( \ce{ZnCl2} \).
1° (last, least reactive) — a primary carbocation with one alkyl group is the least stable.
This is the basis of the Lucas test: tertiary alcohols give turbidity immediately, while primary alcohols give none at room temperature.
NCERT (2026-27 edition):
From Alcohols – “… The reactions of primary and secondary alcohols with HCl require the presence of a catalyst, ZnCl2. With tertiary alcohols, the reaction is conducted by simply shaking the alcohol with concentrated HCl at room temperature. Constant boiling with HBr (48%) is used for preparing alkyl bromide. Good yields of R—I may be obtained by heating alcohols with sodium or potassium iodide in 95% orthophosphoric acid. The order of reactivity of alcohols with a given haloacid is 3°>2°>1°.”
Reaction with Hydrogen Halides – …
